 Hello and welcome to the session I am Deepika here. Let's discuss a question which says a toy company manufactures two types of dolls A and B. Markets and available resources have indicated that the combined production level should not exceed 1200 dolls per B and the one for dolls of type B is at most half of that for dolls of type A. Further the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of rupees 12 and rupees 16 per doll respectively on dolls A and B how many of each should be produced weekly in order to maximize the profit. So let's start the solution according to the question a toy company manufactures two types of dolls A and B. Now we have to find how many of each should be produced weekly in order to maximize the profit. Let send file be the number of dolls A and B produced respectively. So obviously x is greater than equal to 0 and y is greater than equal to 0. Again according to the question the combined production level should not exceed 1200 dolls. Now according to the question the combined production level should not exceed 1200 dolls. So this implies x plus y less than equal to 1200. So this is a constraint on the production of dolls. Again we are given the demand for dolls of type B is at most half of that for dolls of type A. Since the demand for dolls of type B is at most dolls of type A, y is less than equal to 1 over 2x. Again this implies x minus 2y greater than equal to 0. Further we are given the production level of dolls of type A can exceed 3 times the production of dolls of type B by at most 600 units. Since the production level can exceed 3 times the production of dolls of other type by at most 600 units this implies x minus 3y less than equal to 600. Again we are given the company makes profit of rupees 12 and rupees 16 per doll respectively on dolls A and B. Now the total profit in rupees is equal to 12x plus 16y let z is equal to 12x plus 16y. Hence the mathematical formulation of the given problem is maximize z is equal to 12x plus 16y subject to the constraints x plus y less than equal to 1200 let us take this as number 1. x minus 2y greater than equal to 0 let us take this as number 2. x minus 3y less than equal to 600 let us take this as number 3 and x greater than equal to 0 and y greater than equal to 0 let us take this as number 4. So z is equal to 12x plus 16y is our objective function. We have to maximize subject to these given constraints so we will now draw the graph and find the feasible reasons subject to these given constraints. Now the equation corresponding to the inequality 1 that is to the inequality x plus y less than equal to 1200 is x plus y is equal to 1200. Now clearly the points 0 1200 and 1200 0 satisfy the equation x plus y is equal to 1200. So the graph of the line x plus y is equal to 1200 can be drawn by plotting points 0 1200 and 1200 0 and then joining them. Now this is the point 0 1200 and this is the point 1200 0 this line represents our equation x plus y is equal to 1200. Now this 9 divides the plane into 2 half planes so we will consider the half plane which will satisfy 1. Clearly the origin satisfy this inequality so the half plane containing the origin is the graph of 1. Now again the equation corresponding to the inequality x minus 2y greater than equal to 0 is x minus 2y is equal to 0. Now we will plot the points 0 0 and 800 400 on the same graph as this satisfies the equation x minus 2y is equal to 0. That is to get the graph of the equation x minus 2y is equal to 0 we will plot the points 0 0 and 800 400 and then we will join them. Now this is the point 0 0 that is this is the origin and this is the point 800 400 now this line represents our equation x minus 2y is equal to 0. Now again this line divides the plane into 2 half planes so we will consider the half plane which will satisfy 2. Now if we take any point to the right of this line or below this line it will satisfy the inequality 2 to the right of this line is the graph of 2. Again the equation corresponding to the inequality x minus 3y less than equal to 600 is x minus 3y is equal to 600. Now clearly the points 600 0 and 100 200 satisfy the equation x minus 3y is equal to 600. So we will plot these points on the same graph and then we will join them. Now this is the point 600 0 and 100 200 in this line divides the plane into 2 half planes so we will consider the half plane which will satisfy 3. Now if we take any point above this blue line it will satisfy the inequality 3 so the half plane above this line that is above the blue line is the graph of 3. Now x greater than equal to 0 and y greater than equal to 0 implies that the graph flies in the first quadrant only. Hence the shaded region in this graph is the feasible region according to the given constraints. Clearly the shaded region is bounded now we find out the coordinates of its corner points. Now let us take this as a point a so the coordinates of a are 600 0. Now let us take this point as b. Now b is the point of intersection of the line x plus y is equal to 1200 and x minus 3y is equal to 600. Now here from the graph we observe that the coordinates of b are 1050 150 this as the point c so the coordinates of c are 800 400. So here the feasible region is counted with coordinates of its corner points as o with coordinates 0 0 a with coordinates 600 0 b with coordinates 1050 150 and c with coordinates 800 400. So according to the corner point vector maximum value of z will occur at any of these points so we will evaluate z which is equal to 12x plus 16y at each of these points. Now at origin z is equal to 12 into 0 plus 16 into 0 which is equal to 0. Now at the point a with coordinates 600 0 z is equal to 12 into 600 plus 16 into 0 which is equal to 7200. Now at the point b with coordinates 1050 150 z is equal to 12 into 1050 plus 16 into 150 and this is equal to 12600 plus 2400 and this is again equal to 15000. Now at the point c with coordinates 800 400 z is equal to 12 into 800 plus 16 into 400 and this is equal to 9600 6400 and this is again equal to 16000. Maximum value z is equal to 16000 which occurs when x is equal to 800 and y is equal to 400. Thus maximum profit is rupees 16000 which occurs when 800 dollars of type A and 200 dollars of type B are produced. So the answer for the question is that the company should produce 800 dollars of type A and 400 dollars of type B to maximize its profit and then the maximum profit will be rupees 16000. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.